Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A parallel-plate air capacitor of capacitance has a charge of magnitude on each plate. The plates are apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric field magnitude between the plates? (d) What is the surface charge density on each plate?

Short Answer

Expert verified

the potential difference between the plates is 10Kv

the area of each plate is

theelectricfiledmagnitudeis

The electric Charge density is

Step by step solution

01

About Parallel plate capacitor 

A parallel plate capacitor is an arrangement of two metal plates connected in parallel separated from each other by some distance. A dielectric medium occupies the gap between the plates

02

Determine the formula used

,,,

03

determine the potential difference between the plates

Therefore the potential difference between the plates is 10Kv

04

Determine the Area of each plate 

Therefore the area of each plate is

05

:Determine the Electric field magnitude and surface charge density

Therefore theelectricfieldmagnitudeis

The electric Charge density is

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Questions: When a thunderstorm is approaching, sailors at sea sometimes observe a phenomenon called “St. Elmo’s fire,” a bluish flickering light at the tips of masts. What causes this? Why does it occur at the tips of masts? Why is the effect most pronounced when the masts are wet? (Hint: Seawater is a good conductor of electricity.)

Copper has 8.5×1022free electrons per cubic meter. A 71.0-cm

length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 A of

current. (a) How much time does it take for an electron to travel the length

of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 mm)

of the same length that carries the same current. (c) Generally speaking,

how does changing the diameter of a wire that carries a given amount of

current affect the drift velocity of the electrons in the wire?

In a cyclotron, the orbital radius of protons with energy 300keVis 16.0cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300keV. An alpha particle has chargeq=+2e and mass m=6.64×10-27kg. If the magnetic filed isn't changed, what will be the orbital radius of the alpha particles?

When a resistor with resistance Ris connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) (a) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that Rremains constant when the power consumption changes. (b) The resistor is connected to a battery and consumes 5.00 W. What is the voltage of this battery?

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free